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posted in Single-Card Discussion read more

@Boogdish Split cards count for only one.

708.4b The mana cost of a split card is the combined mana costs of its two halves. A split card’s
colors and converted mana cost are determined from its combined mana cost.

posted in Single-Card Discussion read more

It is worth noting this card is also a "key" so the combo "vault-key" can keep is name.

posted in Single-Card Discussion read more

Rule 121.6.

Some spells and abilities refer to counters being put on an object. This refers to putting counters
on that object while it’s on the battlefield and also to an object that’s given counters as it enters the
battlefield.

posted in Single-Card Discussion read more

And a new preview saison is starting.

0_1560787099381_0d707b25-aea2-4852-8a69-da6e3b4e8ea9-image.png

Source: Andrea Mengucci

Scheming Symmetry B

Sorcery
Choose two target players. Each of them searches their library for a card, then shuffles their library and puts that card on top of it.

If you do nothing the opponent will draw his card first but this can be negated by shuffling or extra draw on our turn. Can this card see some play when Imperial seal see almost 0 play ?

posted in Single-Card Discussion read more

@Khahan Unfortunately probability is rarely intuitive.

The mathematical proof:

We consider 3 type of card a, b, c and a deck with Na + Nb + Nc = 60 copies of each card.
The probably of having exactly A + B + C = 7 in a hand of 7 is given by

P(A, B, C; 60) = comb(Na, A) * comb(Nb, B) * comb(Nc, C) / comb(60, 7)

By introducing comb(60 - Na, 7 - A) in the middle we obtain

P(A, B, C; 60) = [comb(Na, A) * comb(60 - Na, 7 - A) / comb(60, 7)] * [comb(Nb, B) * comb(Nc, C) / comb(60 - Na, 7 - A)]

We recognize the probability for 2 type of card and use C = 7 - A - B to obtain

P(A, B, 7 - A - B; 60) = P(A, 7 - A; 60) * P(B, 7 - A - B, 60 - A)

Wikipedia for more detail.

The interpretation is: For each new type of card you have to remove the all the copy of the previous type of card.

This is not easy, I had to redo the proof to convince myself there was an error in the spreadsheet.

posted in Single-Card Discussion read more

@ChubbyRain Nice work.

I think there is a error in "Chance of Green card out of the remaining six cards", your results correspond to a population of 59 = (60 - 1 * Force) but it should be done with population of 56 = (60 - 4 * Force).

Everything else seems good for me.

posted in Vintage Community read more

@joshuabrooks I combined the list of edition of MTG Wiki with the data base of card of MTGJSON (used by Cockatrice), Un-set are excluded. I got a total of 18909 cards.

The number of set (with at least one new card) by year:

0_1559949586826_numberofset.png

The number of new printing by year:

0_1559949590143_numberofcards.png

The year 2019 is ongoing so the number are incomplete.
There seem to have an increase in the number of new printing in the last few years.

posted in Vintage Strategy read more

They can play Spark Double so it is not completely absurd sarcasm. My bad.

posted in Vintage Strategy read more

@Shopsaholic It will be hard to hit four Karns with The Elderspell when there is only three versions of legendary Karn planewalkers.

The probme is that paying BB to destroy planewalkers when most of them cost 3 or 4 and have already use at least one of their abilities is a bit slow. And when they do not play planewalker, its a completely dead card.

posted in Vintage Strategy read more

There is 3 relevant challenges, and in the last one WAR is legal. April 13, April 20 and April 27.